ART. 256]
RECTILINEAR FIGURES.
131
angle dd having its proper sign according as the radial angle 0 is increasing or decreasing when Q passes over each element of the perimeter.
When the perpendicular PN falls within the lamina, the limits of B are 0 and 27T, the expression for the potential is then $Rd$ - 2?r^. When the perpendicular falls outside the lamina the upper and lower limits of 6 are the same, so that fed6=Q and the expression for the potential is simply \Bdd.
256. We may put the expression just found for the potential into another form which is sometimes more useful.
If rdddr is any element of the area of the triangle NQQ', u its distance from P and 0 the angle u makes with the normal to the plane, the solid angle da subtended at P by the triangle is rdddr
—=-0080=1
the limits of u being £ and E.
The potential of the triangular area NQQ' at P is, by Art. 255, equal to
In fig. 1, the perpendicular PN falls within the attracting area. We then find, by integrating all round the perimeter of the area, that the potential at P is
where w is now the solid angle subtended at P by the area.
P P
N
Fig. 1.
Fig. 2.
In fig. 2, the perpendicular PN falls without the area. In this case we must subtract from the potential of NQQf that of NSS'. Since dd is positive for QQr and negative for S'S when a point travels round the curve in the positive direction, the form of the result is unaltered.
Let ds be the length of any elementary arc QQ' of the perimeter, p the perpendicular from N on the tangent at Q. Then since r*d8=pdst the potential at P of
the area takes the form V = I -^- - fw, where the integration extends all round the perimeter, and o> is the solid angle subtended by the lamina at P.
Ex. If the law of force be the inverse fifth power of the distance, show that
the potential of a plane lamina of unit density at a point P is —-5 / ^r, where
8i J •*>'
the integration extends all round the perimeter and the letters have the same meaning as in Art. 255.